cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A364329 G.f. satisfies A(x) = (1 + x^3) * (1 + x*A(x)^2).

Original entry on oeis.org

1, 1, 2, 6, 17, 52, 167, 558, 1912, 6683, 23736, 85426, 310861, 1141837, 4227938, 15764474, 59140089, 223062670, 845388258, 3217750229, 12295043520, 47144444476, 181349473833, 699629022954, 2706327445312, 10494497061015, 40787775234746, 158859378070721
Offset: 0

Views

Author

Seiichi Manyama, Jul 18 2023

Keywords

Crossrefs

Programs

  • Maple
    A364329 := proc(n)
        add( binomial(2*n-6*k+1,k) * binomial(2*n-6*k+1,n-3*k)/(2*n-6*k+1),k=0..n/3) ;
    end proc:
    seq(A364329(n),n=0..70); # R. J. Mathar, Jul 25 2023
  • Mathematica
    nmax = 27; A[_] = 1;
    Do[A[x_] = (1 + x^3)*(1 + x*A[x]^2) + O[x]^(nmax+1) // Normal, {nmax+1}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Mar 03 2024 *)
  • PARI
    a(n) = sum(k=0, n\3, binomial(2*n-6*k+1, k)*binomial(2*n-6*k+1, n-3*k)/(2*n-6*k+1));

Formula

G.f.: A(x) = 2*(1 + x^3) / (1 + sqrt(1-4*x*(1 + x^3)^2)).
a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-6*k+1,k) * binomial(2*n-6*k+1,n-3*k) / (2*n-6*k+1).
D-finite with recurrence (n+1)*a(n) +2*(-2*n+1)*a(n-1) +(n+1)*a(n-3) +6*(-2*n+7)*a(n-4) +6*(-2*n+13)*a(n-7) +2*(-2*n+19)*a(n-10)=0. - R. J. Mathar, Jul 25 2023

A379160 G.f. A(x) satisfies A(x) = (1 + x^4) * (1 + x*A(x)^3).

Original entry on oeis.org

1, 1, 3, 12, 56, 277, 1449, 7872, 43978, 251049, 1457910, 8585724, 51152973, 307770236, 1867362978, 11412645576, 70194232585, 434157164246, 2698676669337, 16849393604484, 105621663128839, 664493496025485, 4194246946152171, 26553277765995984, 168567628890184974, 1072818569798746567
Offset: 0

Views

Author

Seiichi Manyama, Dec 17 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(3*n-12*k+1, k)*binomial(3*n-12*k+1, n-4*k)/(3*n-12*k+1));

Formula

a(n) = Sum_{k=0..floor(n/4)} binomial(3*n-12*k+1,k) * binomial(3*n-12*k+1,n-4*k)/(3*n-12*k+1).
Showing 1-2 of 2 results.